Motivation.

Review of key ideas and equations from the fluid dynamics portion of the class.

Vector displacements.

Those portions of the theory of elasticity that we did cover have the appearance of providing some logical context for the derivation of the Navier-Stokes equation. Our starting point is almost identical, but we now look at displacements that vary with time, forming

We compute a first order Taylor expansion of this differential, defining a symmetric strain and antisymmetric vorticity tensor

\begin{subequations}

\end{subequations}

Allowing us to write

We introduced vector and dual vector forms of the vorticity tensor with

\begin{subequations}

\end{subequations}

or

\begin{subequations}

\end{subequations}

We were then able to put our displacement differential into a partial vector form

Relative change in volume

We are able to identity the divergence of the displacement as the relative change in volume per unit time in terms of the strain tensor trace (in the basis for which the strain is diagonal at a given point)

Conservation of mass

Utilizing Green’s theorem we argued that

We were able to relate this to rate of change of density, computing

An important consequence of this is that for incompressible fluids (the only types of fluids considered in this course) the divergence of the displacement .

Constitutive relation

We consider only Newtonian fluids, for which the stress is linearly related to the strain. We will model fluids as disjoint sets of hydrostatic materials for which the constitutive relation was previously found to be

Conservation of momentum (Navier-Stokes)

As in elasticity, our momentum conservation equation had the form

where are the components of the external (body) forces per unit volume acting on the fluid.

Utilizing the constitutive relation and explicitly evaluating the stress tensor divergence we find

Since we treat only incompressible fluids in this course we can decompose this into a pair of equations

\begin{subequations}

\end{subequations}

No slip condition

We’ll find in general that we have to solve for our boundary value conditions. One of the important constraints that we have to do so will be a requirement (experimentally motivated) that our velocities match at an interface. This was illustrated with a rocker tank video in class.

This is the no-slip condition, and includes a requirement that the fluid velocity at the boundary of a non-moving surface is zero, and that the fluid velocity on the boundary of a moving surface matches the rate of the surface itself.

For fluids and separated at an interface with unit normal and unit tangent we wrote the no-slip condition as

\begin{subequations}

\end{subequations}

For the problems we attempt, it will often be enough to consider only the tangential component of the velocity.

Traction vector matching at an interface.

As well as matching velocities, we have a force balance requirement at any interface. This will be expressed in terms of the traction vector

where is the normal pointing from the interface into the fluid (so the traction vector represents the force of the interface on the fluid). When that interface is another fluid, we are able to calculate the force of one fluid on the other.

In addition the the constraints provided by the no-slip condition, we’ll often have to constrain our solutions according to the equality of the tangential components of the traction vector

We’ll sometimes also have to consider, especially when solving for the pressure, the force balance for the normal component of the traction vector at the interface too

As well as having a messy non-linear PDE to start with, our boundary value constraints can be very complicated, making the subject rich and tricky.

Flux

A number of problems we did asked for the flux rate. A slightly more sensible physical quantity is the mass flux, which adds the density into the mix

Worked problems from class.

A number of problems were tackled in class

channel flow with external pressure gradient

shear flow

pipe (Poiseuille) flow

steady state gravity driven film flow down a slope.

Channel and shear flow

An example that shows many of the features of the above problems is rectilinear flow problem with a pressure gradient and shearing surface. As a review let’s consider fluid flowing between surfaces at , the lower surface moving at velocity and pressure gradient we find that Navier-Stokes for an assumed flow of takes the form

We find that this reduces to

with solution

Application of the no-slip velocity matching constraint gives us in short order

With this is the channel flow solution, and with this is the shearing flow solution.

Having solved for the velocity at any height, we can also solve for the mass or volume flux through a slice of the channel. For the mass flux per unit time (given volume flux )

we find

We can also calculate the force of the boundaries on the fluid. For example, the force per unit volume of the boundary at on the fluid is found by calculating the tangential component of the traction vector taken with normal . That tangent vector is found to be

The tangential component is the component evaluated at , so for the lower and upper interfaces we have

so the force per unit area that the boundary applies to the fluid is

Does the sign of the velocity term make sense? Let’s consider the case where we have a zero pressure gradient and look at the lower interface. This is the force of the interface on the fluid, so the force of the fluid on the interface would have the opposite sign

This does seem reasonable. Our fluid flowing along with a positive velocity is imparting a force on what it is flowing over in the same direction.

Hydrostatics.

We covered hydrostatics as a separate topic, where it was argued that the pressure in a fluid, given atmospheric pressure and height from the surface was

As noted below in the surface tension problem, this is also a consequence of Navier-Stokes for (following from ).

We noted that replacing the a mass of water with something of equal density would not change the non-dynamics of the situation. We then went on to define Buoyancy force, the difference in weight of the equivalent volume of fluid and the weight of the object.

Mass conservation through apertures.

It was noted that mass conservation provides a relationship between the flow rates through apertures in a closed pipe, since we must have

and therefore for incompressible fluids

So if we must have .

Curve for tap discharge.

We can use this to get a rough idea what the curve for water coming out a tap would be. Suppose we measure the volume flux, putting a measuring cup under the tap, and timing how long it takes to fill up. We then measure the radii at different points. This can be done from a photo as in figure (1).

Figure 1: Tap flow measurement.

After making the measurement, we can get an idea of the velocity between two points given a velocity estimate at a point higher in the discharge. For a plain old falling mass, our final velocity at a point measured from where the velocity was originally measured can be found from Newton’s law

Solving for , we find

Mass conservation gives us

or

For the image above I measured a flow rate of about 250 ml in 10 seconds. With that, plus the measured radii at 0 and , I calculated that the average fluid velocity was , vs a free fall rate increase of . Not the best match in the world, but that’s to be expected since the velocity has been considered uniform throughout the stream profile, which would not actually be the case. A proper treatment would also have to treat viscosity and surface tension.

In figure (2) is a plot of the measured radial distance compared to what was computed with 11.41. The blue line is the measured width of the stream as measured, the red is a polynomial curve fitted to the raw data, and the green is the computed curve above.

Figure 2: Comparison of measured stream radii and calculated.

Bernoulli equation.

With the body force specified in gradient for

and utilizing the vector identity

we are able to show that the steady state, irrotational, non-viscous Navier-Stokes equation takes the form

or

This is the Bernoulli equation, and the constants introduce the concept of streamline.

FIXME: I think this could probably be used to get a better idea what the tap stream radius is, than the method used above. Consider the streamline along the outermost surface. That way you don’t have to assume that the flow is at the average velocity rate uniformly throughout the stream. Try this later.

Surface tension.

Surfaces, normals and tangents.

We reviewed basic surface theory, noting that we can parameterize a surface as in the following example

Computing the gradient we find

Recalling that the gradient is normal to the surface we can compute the unit normal and unit tangent vectors

\begin{subequations}

\end{subequations}

Laplace pressure.

We covered some aspects of this topic in class. [1] covers this topic in typical fairly hard to comprehend) detail, but there’s lots of valuable info there. section 2.4.9-2.4.10 of [2] has small section that’s a bit easier to understand, with less detail. Recommended in that text is the “Surface Tension in Fluid Mechanics” movie which can be found on youtube in three parts \youtubehref{DkEhPltiqmo}, \youtubehref{yiixltf\_HKw}, \youtubehref{5d6efCcwkWs}, which is very interesting and entertaining to watch.

It was argued in class that the traction vector differences at the surfaces between a pair of fluids have the form

where is the tangential (interfacial) gradient, is the surface tension, a force per unit length value, and is the radius of curvature.

In static equilibrium where (since if ), then dotting with we must then have

Surface tension gradients.

Considering the tangential component of the traction vector difference we find

If the fluid is static (for example, has none of the creep that we see in the film) then we must have . It’s these gradients that are responsible for capillary flow and other related surface tension driven motion (lots of great examples of that in the film).

Surface tension for a spherical bubble.

In the film above it is pointed out that the surface tension equation we were shown in class

is only for spherical objects that have a single radius of curvature. This formula can in fact be derived with a simple physical argument, stating that the force generated by the surface tension along the equator of a bubble (as in figure (3)), in a fluid would be balanced by the difference in pressure times the area of that equatorial cross section. That is

Figure 3: Spherical bubble in liquid.

Observe that we obtain 13.52 after dividing through by the area.

A sample problem. The meniscus curve.

To get a better feeling for this, let’s look to a worked problem. The most obvious one to try to attempt is the shape of a meniscus of water against a wall. This problem is worked in [1], but it is worth some extra notes. As in the text we’ll work with axis up, and the fluid up against a wall at as illustrated in figure (4).

Figure 4: Curvature of fluid against a wall.

The starting point is a variation of what we have in class

where is the atmospheric pressure, is the fluid pressure, and the (signed!) radius of curvatures positive if pointing into medium 1 (the fluid).

For fluid at rest, Navier-Stokes takes the form

With we have

or

We have , the atmospheric pressure, so our pressure difference is

We have then

One of our axis of curvature directions is directly along the axis so that curvature is zero . We can fix the constant by noting that at , , we have no curvature . This gives

That leaves just the second curvature to determine. For a curve our absolute curvature, according to [3] is

Now we have to fix the sign. I didn’t recall any sort of notion of a signed radius of curvature, but there’s a blurb about it on the curvature article above, including a nice illustration of signed radius of curvatures can be found in this wikipedia radius of curvature figure for a Lemniscate. Following that definition for a curve such as we’d have a positive curvature, but the text explicitly points out that the curvatures are will be set positive if pointing into the medium. For us to point the normal into the medium as in the figure, we have to invert the sign, so our equation to solve for is given by

The text introduces the capillary constant

Using that capillary constant to tidy up a bit and multiplying by a integrating factor we have

we can integrate to find

Again for we have , , so . Rearranging we have

Integrating this with Mathematica I get

It looks like the constant would have to be fixed numerically. We require at

but we don’t have an explicit function for .

Non-dimensionality and scaling.

With the variable transformations

we can put Navier-Stokes in dimensionless form

Here is Reynold’s number

A relatively high or low Reynold’s number will effect whether viscous or inertial effects dominate

The importance of examining where one of these effects can dominate was clear in the Blassius problem, where doing so allowed for an analytic solution that would not have been possible otherwise.

The aim of this appears to be as an illustration that the boundary layer thickness grows with .

FIXME: really need to plot 16.80.

Oscillatory flow.

Another worked problem in the boundary layer topic was the Stokes boundary layer problem with a driving interface of the form

with an assumed solution of the form

we found

\begin{subequations}

\end{subequations}

This was a bit more obvious as a boundary layer illustration since we see the exponential drop off with every distance multiple of .

Blassius problem (boundary layer thickness in flow over plate).

We examined the scaling off all the terms in the Navier-Stokes equations given a velocity scale , vertical length scale and horizontal length scale . This, and the application of Bernoulli’s theorem allowed us to make construct an approximation for Navier-Stokes in the boundary layer

\begin{subequations}

\end{subequations}

With boundary conditions

With a similarity variable

and stream functions

and

we were able to show that our velocity dependence was given by the solutions of

This was done much more clearly in [4] and I worked this problem myself with a hybrid approach (non-dimensionalising as done in class).

FIXME: the end result of this is a plot (a nice one can be found in [5]). That plot ends up being one that’s done in terms of the similarity variable . It’s not clear to me how this translates into an actual velocity profile. Should plot these out myself to get a feel for things.

Singular perturbation theory.

The non-dimensional form of Navier-Stokes had the form

where the inverse of Reynold’s number

can potentially get very small. That introduces an ill-conditioning into the problems that can make life more interesting.

We looked at a couple of simple LDE systems that had this sort of ill conditioning. One of them was

for which the exact solution was found to be

The rough idea is that we can look in the domain where and far from that. In this example, with far from the origin we have roughly

so we have an asymptotic solution close to . Closer to the origin where we can introduce a rescaling to find

This gives us

for which we find

Stability.

We characterized stability in terms of displacements writing

and defining

Oscillatory unstability. .

Marginal unstability. .

Neutral stability. .

Thermal stability: Rayleigh-Benard problem.

We considered the Rayleigh-Benard problem, looking at thermal effects in a cavity. Assuming perturbations of the form

and introducing an equation for the base state

we found

Operating on this with we find

from which we apply back to 18.103 and take just the z component to find

With an assumption that density change and temperature are linearly related

and operating with the Laplacian we end up with a relation that follows from the momentum balance equation

where we’ve introduced the Rayleigh number and Prandtl number’s
\begin{subequations}

\end{subequations}

We were able to construct some approximate solutions for a problem similar to these equations using an assumed solution form

Using these we are able to show that our PDEs are similar to that of

where . Using the trig solutions that fall out of this we were able to find the constraint

which for , this gives us the critical value for the Rayleigh number

which is the boundary for thermal stability or instability.

The end result was a lot of manipulation for which we didn’t do any sort of applied problems. It looks like a theory that requires a lot of study to do anything useful with, so my expectation is that it won’t be covered in detail on the exam. Having some problems to know why we spent two days on it in class would have been nice.

Motivation.

Exersize 6.1 from [1] is to show that the traction vector can be written in vector form (a rather curious thing to have to say) as

Note that the text uses a wedge symbol for the cross product, and I’ve switched to standard notation. I’ve done so because the use of a Geometric-Algebra wedge product also can be used to express this relationship, in which case we would write

In either case we have

(where the primes indicate the scope of the gradient, showing here that we are operating only on , and not ).

After computing this, lets also compute the stress tensor in cylindrical and spherical coordinates (a portion of that is also problem 6.10), something that this allows us to do fairly easily without having to deal with the second order terms that we encountered doing this by computing the difference of squared displacements.

We’ll work primarily with just the strain tensor portion of the traction vector expressions above, calculating

We’ll see that this gives us a nice way to interpret these tensor relationships. The interpretation was less clear when we computed this from the second order difference method, but here we see that we are just looking at the components of the force in each of the respective directions, dependent on which way our normal is specified.

Verifying the relationship.

Let’s start with the the plain old cross product version

We can also put the double cross product in wedge product form

Equivalently (and easier) we can just expand the dot product of the wedge and the vector using the relationship

We can now move on to compute the directional derivatives and complete the strain calculation in cylindrical coordinates. Let’s consider this computation of the stress for normals in each direction in term.

With .

Our directional derivative component for a normal direction doesn’t have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

With .

Our directional derivative component for a normal direction will have some cross terms since both and are functions of

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

With .

Like the normal direction, our directional derivative component for a normal direction will not have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

Summary.

\begin{subequations}

\end{subequations}

Spherical strain tensor.

Having done a first order cylindrical derivation of the strain tensor, let’s also do the spherical case for completeness. Would this have much utility in fluids? Perhaps for flow over a spherical barrier?

We need the gradient in spherical coordinates. Recall that our spherical coordinate velocity was

and our gradient mirrors this structure

We also previously calculated \inbookref{phy454:continuumL2}{eqn:continuumL2:1010} the unit vector differentials

\begin{subequations}

\end{subequations}

and can use those to read off the partials of all the unit vectors

Finally, our velocity in spherical coordinates is just

from which we can now compute the curl, and the directional derivative. Starting with the curl we have

So we have

With .

The directional derivative portion of our strain is

The other portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

This is consistent with (15.20) from [3] (after adjusting for minor notational differences).

With .

Now let’s do the direction. The directional derivative portion of our strain will be a bit more work to compute because we have variation of the unit vectors

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

This again is consistent with (15.20) from [3].

With .

Finally, let’s do the direction. This directional derivative portion of our strain will also be a bit more work to compute because we have variation of the unit vectors

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Motivation.

In an informal discussion after class, it was claimed that the steady state flow down a plane would have constant height, unless you bring surface tension effects into the mix. Part of that statement just doesn’t make sense to me. Consider the forces acting on the fluid in the figure (\ref{fig:inclinedFlowWithoutConstantHeightAssumption:inclinedFlowWithoutConstantHeightAssumptionFig1})

In the inclined reference frame we have a component of the force acting downwards (in the negative y-axis direction), and have a component directed down the x-axis. Wouldn’t this act to both push the fluid down the plane and push part of the fluid downwards? I’d expect this to introduce some vorticity as depicted.

While we are just about to start covered surface tension, perhaps this is just allowing the surface to vary, and then solving the Navier-Stokes equations that result. Let’s try setting up the Navier-Stokes equation for steady state viscous flow down a plane without any assumption that the height is constant and see how far we can get.

The equations of motion.

We’ll use the same coordinates as before, with the directions given as in figure (\ref{fig:inclinedFlowWithoutConstantHeightAssumption:inclinedFlowWithoutConstantHeightAssumptionFig2}). However, this time, we let the height of the fluid at any distance down the plane from the initial point vary.

For viscous incompressible flow down the plane our equations of motion are

\begin{subequations}

\end{subequations}

Now, can we kill the time dependent term? Even allowing for to vary with and introducing a non-horizontal flow component, I think that we can. If the flow at is constant, not varying at all with time, I think it makes sense that we will have no time dependence in the flow for . So, I believe that our starting point is as above, but with the time derivatives killed off. That is

\begin{subequations}

\end{subequations}

These don’t look particularily easy to solve, and we haven’t even set up the boundary value constraints yet. Let’s do that next.

The boundary value constraints.

One of out constraints is the no-slip condition for the velocity components at the base of the slope

We should also have a zero tangential component to the traction vector at the interface. We need to consider some geometry, and refer to figure (\ref{fig:inclinedFlowWithoutConstantHeightAssumption:inclinedFlowWithoutConstantHeightAssumptionFig3})

What is the normal component of the traction vector at the interface? We can calculate

So this component of the traction vector is

For the purely recilinear flow, with and , as a a consequence of NS and our assumptions, all but the pressure portion of this component of the traction vector was zero. The force balance equation for the interface was therefore just a matching of the pressure with the external (ie: air) pressure.

In this more general case we have the same thing, but the non-pressure portions of the traction vector are all zero in the medium. Outside of the fluid (in the air say), we have assumed no motion, so this force balance condition becomes

Again assuming no motion of the air, with air pressure , this is

Observe that for the horizontal flow problem, where and , this would have been nothing more than a requirement that , but now that we introduce downwards motion and allow the height to vary, the pressure matching condition becomes a much more complex beastie.

Laplacian of Pressure and Vorticity.

Supposing that we are neglecting the non-linear term of the Navier-Stokes equation. For incompressible steady state flow, without any external forces, we would then have

\begin{subequations}

\end{subequations}

How do we actually solve this beastie?

Separation of variables?

Considering this in 2D, assuming no z-dependence, with we have

Attempting separation of variables seems like something reasonable to try. With

we get

I couldn’t find a way to substitute any of these into the other that would allow me to separate them, but perhaps I wasn’t clever enough.

In terms of vorticity?

The idea of substuting the zero divergence equation will clearly lead to something a bit simpler. Treating the Laplacian as a geometric (clifford) product of two gradients we have

Writing out the vorticity (bivector) in component form, and writing for the 2D pseudoscalar, we have

It seems natural to write

so that Navier-Stokes takes the form

Operating on this from the left with another gradient we find that this combination of pressure and vorticity must satisfy the following multivector Laplacian equation

However, note that is a scalar operator, and this zero identity has both scalar and pure bivector components. Both must separately equal zero

\begin{subequations}

\end{subequations}

Note that we can obtain 4.29 much more directly, if we know that’s what we want to do. Just operate on 4.16a with the gradient from the left right off the bat. We find

Again, we have a multivector equation scalar and bivector parts, that must separately equal zero. With the magnitude of the vorticity given by 4.26, we once again obtain 4.29. This can be done in plain old vector algebra as well by operating on the left not by the gradient, but separately with a divergence and curl operator.

Pressure and vorticity equations with the non-linear term retained.

If we add back in our body force, and assume that it is constant (i.e. gravity), then this this will get killed with the application of the gradient. We’ll still end up with one Laplacian for pressure, and one for vorticity. That’s not the case for the inertial term of Navier-Stokes. Let’s take the divergence and curl of this and see how we have to modify the Laplacian equations above.

Starting with the divergence, with summation implied over repeated indexes, we have

So we have

We are working with the incompressibility assumption so we kill off the last term. Our velocity ends up introducing a non-homogeneous forcing term to the Laplacian pressure equation and we’ve got something trickier to solve

Now let’s see how our vorticity Laplacian needs to be modified. Taking the curl of the implusive term we have

So we have

Putting things back together, our vorticity equation is

Or, with

this is

It is this and 4.31 that we really have to solve. Before moving on, let’s write out all the non-boundary condition equations in coordinate form for the 2D case that we are interested in here. We have

\begin{subequations}

\end{subequations}

Our solution has to satisfy these equations, as well as still satisfying the original Navier-Stokes system 2.2 that includes our gravitational term, and also has to satisfy both of our boundary value constraints 3.12, 3.15, and have . Wow, what a mess! And this is all just the steady state problem. Imagine adding time into the mix too!

Now what?

The strategy that I’d thought to attempt to tackle a problem with is something along the following lines

\begin{itemize}
\item First ignore the non-linear terms. Find solutions for the homogenous vorticity and pressure Laplacian equations that satisfy our boundary value conditions, and use that to find a first solution for .
\item Use this to solve for and from the vorticity.
\end{itemize}

However, after having gotten this far, just setting up the problem for solution, I think that it’s perhaps best not to try to solve it yet, and study some more first. I have a feeling that there are likely a number of techniques that have been developed that will likely be useful to know before I try to plow my way through things. It’s interesting to see just how tricky the equations of motion become when one doesn’t make unrealistic assumptions. I have a feeling that to actually attempt this specific problem, I may very well need a computer and numerical techniques.

Motivation.

I didn’t manage my time well enough on the midterm to complete it (and also missed one easy part of the second question). For later review purposes, here is either what I answered, or what I think I should have answered for these questions.

Problem 1.

-waves, -waves, and Love-waves.

\begin{itemize}
\item Show that in -waves the divergence of the displacement vector represents a measure of the relative change in the volume of the body.

Answer.
The -wave equation was a result of operating on the displacement equation with the divergence operator

we obtain

We have a wave equation where the “waving” quantity is . Explicitly

Recall that, in a coordinate basis for which the strain is diagonal we have

Expanding in Taylor series to we have for (no sum)

so the displaced volume is

Since

We have

or

The relative change in volume can therefore be expressed as the divergence of , the displacement vector, and it is this relative volume change that is “waving” in the -wave equation as illustrated in the following (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig1}) sample 1D compression wave

\item Between a -wave and an -wave which one is longitudinal and which one is transverse?

Answer.-waves are longitudinal.

-waves are transverse.

\item Whose speed is higher?

Answer.
From the formula sheet we have

so -waves travel faster than -waves.

\item Is Love wave a body wave or a surface wave?

Answer.

Love waves are surface waves, traveling in a medium that can slide on top of another surface. These are characterized by vorticity rotating backwards compared to the direction of propagation as shown in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig2})

(b) constitutive relation, Newtonian fluids, and no-slip conditions.

\begin{itemize}
\item In continuum mechanics what do you mean by \textit{constitutive relation}?Answer. The constitutive relation is the stress-strain relation, generally

for isotropic solids we model this as

and for Newtonian fluids

\item What is the definition of a Non-Newtonian fluid?Answer.

A non-Newtonian fluid would be one with a more general constitutive relationship.

Grading note. I lost a mark here. I think the answer that was being looked for (as in [1]) was that a Newtonian fluid is one with a linear stress strain relationship, and a non-Newtonian fluid would be one with a non-linear relationship. According to [2] an example of a non-Newtonian material that we are all familiar with is Silly Putty. This linearity is also how a Newtonian fluid was defined in the notes, but I didn’t remember that (this isn’t really something we use since we assume all fluids and materials are Newtonian in any calculations that we do).

\item What do you mean by \emph{no-slip} boundary condition at a fluid-fluid interface?

Exam time management note. Somehow in my misguided attempt to be complete, I missed this question amongst the rest of my verbosity).

Answer.
The no slip boundary condition is just one of velocity matching. At a non-moving boundary, the no-slip condition means that we’ll require the fluid to also have no velocity (ie. at that interface the fluid isn’t slipping over the surface). Between two fluids, this is a requirement that the velocities of both fluids match at that point (and all the rest of the points along the region of the interaction.)

\item Write down the continuity equation for an incompressible fluid.Answer.

An incompressible fluid has

but since we also have

A consequence is that for an incompressible fluid. Let’s recall where this statement comes from. Looking at mass conservation, the rate that mass leaves a volume can be expressed as

(the minus sign here signifying that the mass is leaving the volume through the surface, and that we are using an outwards facing normal on the volume.)

If the surface bounding the volume doesn’t change with time (ie. ) we can write

or

so that in differential form we have

Expanding the divergence by chain rule we have

but this is just

So, for an incompressible fluid (one for which ), we must also have .

\end{itemize}

Problem 2.

Statement.

Consider steady simple shearing flow as shown in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFigQ1}) with imposed constant pressure gradient (), being a positive number, of a single layer fluid with viscosity .

The boundary conditions are no-slip at the lower plate (). The top plate is moving with a velocity at and fluid is sticking to it, so , being a positive number. Using the Navier-Stokes equation.

\begin{itemize}
\item Derive the velocity profile of the fluid.

Answer

Our equations of motion are

\begin{subequations}

\end{subequations}

Here, we’ve used the steady state condition and are neglecting gravity, and kill off our mass compression term with the incompressibility assumption. In component form, what we have left is

with , we must have

which leaves us with just

Having dropped the partials we really just want to integrate our very simple ODE a couple times

Integrate once

and once more to find the velocity

Let’s incorporate an additional constant into

so that we have

(I didn’t do use this way on the exam, nor did I include the factor of in the first integration constant, but both of these should simplify the algebra since we’ll be evaluating the boundary value conditions at .)

Applying the velocity matching conditions we have for the lower and upper plates respectively

Adding these we find

and subtracting find

Our velocity is

or rearranged a bit

\item Draw the velocity profile with the direction of the flow of the fluid when , .

\item Draw the velocity profile with the direction of the flow of the fluid when , .

Answer

With , we have a plain old shear flow

This is linear with minimum velocity at , and a maximum of at . This is plotted in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig4}
\caption{Shear flow.}
\end{figure}

\item Using linear superposition draw the velocity profile of the fluid with the direction of flow qualitatively when , . (i) low , (ii) large .Exam time management note. Somehow I missed this question when I wrote the exam … I figured this out right at the end when I’d run out of time by being too verbose elsewhere. I’m really not very good at writing exams in tight time constraints anymore.

Answer

For low we’ll let the parabolic dominate, and can graphically add these two as in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig5}
\caption{Superposition of shear and parabolic flow (low )}
\end{figure}
For high , we’ll let the shear flow dominate, and have plotted this in figure (\ref{fig:continuumMidtermReflection:continuumMidtermReflectionFig6})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumMidtermReflectionFig6}
\caption{Superposition of shear and parabolic flow (high )}
\end{figure}

\item Calculate the maximum speed when , .Answer

Since our acceleration is

our extreme values occur at

At this point, our velocity is

or just

\item Calculate the flux (the volume flow rate) when , .Answer

An element of our volume flux is

Looking at the volume flux through the width is then

\item Calculate the mean speed when , .Answer

Exam time management note. I squandered too much time on other stuff and didn’t get to this part of the problem (which was unfortunately worth a lot). This is how I think it should have been answered.

We’ve done most of the work above, and just have to divide the flux by . That is

\item Calculate the tangential force (per unit width) on the strip of the wall when , .Answer

Our traction vector is

So the directed component of the traction vector is just

We’ve calculated that derivative above in 3.41, so we have

so at we have

To see the contribution of this force on the lower wall over an interval of length we integrate, but this amounts to just multiplying by the length of the segment of the wall

Disclaimer.

This problem set is as yet ungraded.

Problem Q1.

Statement

Imagine a steady rectilinear blood flow of the form through a two dimensional artery. It is driven by a constant pressure gradient maintained by an external `heart’. The top and bottom walls of the artery are distance apart and the fluid satisfies no-slip boundary conditions at the walls. Assuming that the fluid is Newtonian,

\begin{enumerate}
\item Show that the Navier-Stokes equation reduces to

where is the viscosity of the blood.

\item Show that the velocity profile of the fluid inside the artery is a parabolic profile.
\item What is the maximum speed of the fluid? Draw the velocity profile to show where the maximum speed occurs inside the artery.
\item If due to smoking etc., the viscosity of the blood gets doubled, then what should be the new pressure gradient to be maintained by the `heart’ to keep the liquid flux through the artery at the same level as the non-smoking one?
\end{enumerate}

Solution. Part 1. Navier-Stokes equation for the system.

The Navier-Stokes equation, for an incompressible unidirectional fluid , assuming that there is no dependence, takes the form

\begin{subequations}

\end{subequations}

With a steady state assumption we kill the term, and 2.2d kills of the x-component of the Laplacian and our non-linear inertial term on the LHS, leaving just

\begin{subequations}

\end{subequations}

With , we have , so 2.3a is reduced to

Finally, since we have an assumption of no z-dependence () and from the incompressibility assumption 2.2d (), we have

as desired.

Solution. Part 2. Velocity profile.

Integrating 2.5 twice, we have

Application of the no-slip boundary value condition , we have

Adding and subtracting these, we find

\begin{subequations}

\end{subequations}

so the velocity is given by the parabolic function

Solution. Part 3. Maximum speed of the flow.

It is clear that the maximum speed of the fluid is found at

The velocity profile for this flow is drawn in figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig1}).

Solution. Part 4. Effects of viscosity doubling.

With our velocity being dependent on the ratio, it is clear, even without calculating the flux, that we will need twice the pressure gradient if the viscosity is doubled to maintain the same flux through the artery and veins. To demonstrate this more thoroughly we can calculate this mass flux. For an element of mass leaving a portion of the conduit, bounded by the plane normal to we have

where is the magnitude of the pressure gradient before the bad habits kicked in. The smoker’s poor little heart (soon to be a big overworked and weak heart) has to generate pressure gradients that are twice as big to get the same quantity of blood distributed through the body.

Problem Q2.

Statement

Consider steady simple shearing flow with no imposed pressure gradient of a two layer fluid with viscosity

The boundary conditions are no-slip at the lower plate and at . The top plate is moving with a velocity at and fluid is sticking to it. using the continuity of tangential (shear) stress at the interface ()

\begin{itemize}
\item Derive the velocity profile of the two fluids.
\item Calculate the maximum speed.
\item Calculate the mean speed.
\item Calculate the flux (the volume flow rate.)
\item Calculate the tangential force (per unit width) on the strip of the wall .
\item Calculate the tangential force (per unit width) on the strip at the interface by the top fluid on the lower fluid.
\end{itemize}

Part 1. Derive the velocity profile of the two fluids.

With coordinates referring to figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig2}), our steady flow for layers and has the form

as we found in Q1. Only the boundary value conditions and the driving pressure are different here. In this problem and the next, we have constant pressure gradients to deal with, so we really have just the pair of equations

to solve. For this Q2 problem we have , so the algebra to match our boundary value constraints becomes a bit easier. Our boundary value constraints are

\begin{subequations}

\end{subequations}

plus one more to match the tangential components of the traction vector with respect to the normal . The components of that traction vector are

but we are only interested in the horizontal component which is

So the matching the tangential components of the traction vector at the interface gives us our last boundary value constraint

and we are ready to do our remaining bits of algebra. We wish to solve the pair of equations

for the four integration constants and using our boundary value constraints. The linear system to solve is

With , we have

Subtracting these to solve for we find

This gives us everything we need

Referring back to 3.20 our velocities are

Checking, we see at a glance we see that we have , , , and as desired.

As an example, let’s add some numbers. With mercury and water in layers and respectively, we have

This is plotted with in figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumProblemSet2Fig3}
\caption{Two layer shearing flow with water over mercury.}
\end{figure}

Part 2. Calculate the maximum speed.

With , and linearly decreasing, then linearly decreasing further from the value at , it is clear that the maximum speed, no matter the viscosities of the fluids, is on the upper moving interface. This maximum takes the value .

Part 3. Calculate the mean speed.

As linear functions the average speeds of the respective fluids fall on the midpoints . These are

Averaging these two gives us the overall average, so we find

Part 4. Calculate the flux

We can calculate the volume flux, much like the mass flux (although the mass flux seems a more sensible quantity to calculate). Looking at the rate of change of an element of fluid passing through the plane we have

Integrating over the total height, for a width we have

So our volume flux through a width is

Part 5. Tangential force on the wall.

From 3.32 the tangential components of our traction vectors are

and

We see that the tangential component of the traction vector is a constant throughout both fluids. Allowing this force to act on a length of the lower wall, our force per unit width over that strip is just

The negative value here makes sense since it is acting to push the fluid backwards in the direction of the upper wall motion.

Part 6. Tangential force on the lower fluid.

Due to constant nature of the tangential component of the traction vector shown above, the force per unit width of the upper fluid acting on the lower fluid, is also given by 3.41.

Problem Q3.

Statement

If on top of the problem described above a constant pressure gradient is applied between the boundaries , describe qualitatively what type of flow profile you would expect in the steady state. Draw the velocity profiles for two cases (i) (ii) . Explain your result.

Solution

We showed earlier that the Navier-Stokes equations for this Q3 case, where is non-zero were given by 3.17, which restated is

Our solutions will now necessarily be parabolic, of the form

with the tangential traction vector components given by

The boundary value constants become a bit messier to solve for, and should we wish to do so we’d have to solve the system

Without actually solving this system we should expect that our solution will have the form of our pure shear flow, with parabolas superimposed on these linear flows. For a higher viscosity bottom layer , this should look something like figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig4}) whereas for the higher viscosity on the top, these would be roughly flipped as in figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig5}).

we have a discontinuous jump in density. We will have to consider three boundary value constraints

\begin{enumerate}
\item Mass balance. This is the continuity equation.
\item Momentum balance. This is the Navier-Stokes equation.
\item Energy balance. This is the heat equation.
\end{enumerate}

We have not yet discussed the heat equation, but this is required for non-isothermal problems.

The boundary condition at the interface is given by the stress balance. Denoting the difference in the traction vector at the interface by

Here the gradient is in the tangential direction of the surface as in figure (\ref{fig:continuumL15:continuumL15Fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig2}
\caption{normal and tangent vectors on a curve.}
\end{figure}

In the normal direction

With the traction vector having the value

We have in the normal direction

With on the surface, and we have

Returning to we have

This is the Laplace pressure. Note that the sign of the difference is significant, since it effects the direction of the curvature. This is depicted pictorially in figure (\ref{fig:continuumL15:continuumL15Fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig3}
\caption{pressure and curvature relationships}
\end{figure}

Question In [1] the curvature term is written

Why the difference? Answer: This is to account for non-spherical surfaces, with curvature in two directions. Illustrating by example, imagine a surface like as in figure (\ref{fig:continuumL15:continuumL15Figq})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Figq}
\caption{Example of non-spherical curvature.}
\end{figure}

Followup required to truly understand things

While, this review clarifies things, we still don’t really know how the surface tension term is defined. Nor have we been given any sort of derivation of 2.1, from which the end result follows.

I’m assuming that is a property of the two fluids at the interface, so if you have, for example, oil and vinegar in a bottle, we have surface tension and curvature that’s probably related to how the two of these interact. If there is still a mixing or settling process occurring, I’d imagine that this could even vary from point to point on the surface (imagine adding soap to a surface where stuff can float until the soap mixes in enough that things start sinking in the radius of influence of the soap).

Surface tension gradients.

Now consider the tangential component of the traction vector

So we see that for a static fluid, we must have

For a static interface there cannot be any surface tension gradient. This becomes very important when considering stability issues. We can have surface tension induced flow called capillary, or mandarin (?) flow.

Reynold’s number.

In Navier-Stokes after making non-dimensionalization changes of the form

the control parameter is like Reynold’s number.

In NS

We call the term

the inertial term. It is non-zero only when something is being “carried along with the velocity”. Consider a volume fixed in space and one that is moving along with the fluid as in figure (\ref{fig:continuumL15:continuumL15Fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig4}
\caption{Moving and fixed frame control volumes in a fluid.}
\end{figure}

All of our viscosity dependence shows up in the Laplacian term, so we can roughly characterize the Reynold’s number as the ratio

In figures (\ref{fig:continuumL15:continuumL15Fig5}), (\ref{fig:continuumL15:continuumL15Fig6}) we have two illustrations of viscous and non-viscous regions the first with a moving probe pushing its way through a surface, and the second with a wing set at an angle of attack that generates some turbulence. Both are illustrations of the viscous and inviscous regions for the two flows. Both of these are characterized by the Reynold’s number in some way not really specified in class. One of the points of mentioning this is that when we are in an essentially inviscous region, we can neglect the viscosity () term of the flow.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig5}
\caption{continuumL15Fig5}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig6}
\caption{continuumL15Fig6}
\end{figure}

and the suffix and prefix indicates that we are considering the interface between fluids labelled and (liquid and air respectively in the diagram).

For a derivation see Prof after class?

Force balance along the normal direction gives

If you do this calculation, you will get

I think this was called the Laplace equation?

Question: How was defined? A: Energy per unit area.

Figure (\ref{fig:continuumL14:continuumL14fig4}) was given as part of an explaination of surface tension and curvature, but I missed part of that discussion. Perhaps this is elaborated on in the class notes?

Non dimensionalization and scaling

Motivation.

By scaling we mean how much detail do you want to look at in the analysis. Consider the figure (\ref{fig:continuumL14:continuumL14fig5a}) where we imagine that we zoom in on something that appears smooth from a distance, but perhaps grandular close up as in figure (\ref{fig:continuumL14:continuumL14fig5b}). Picking the length scale to be used in this case can be very important.

Disclaimer.

Navier-Stokes equation.

The Navier-Stokes equation (our fluids equivalent to Newton’s second law) was found to be

In this course we’ll focus on the incompressible case where we have

We watched a video of the rocking tank as in figure (\ref{fig:continuumL11:continuumL11fig1}). The boundary condition that accounted for the matching of the die marker is that we have no slipping at the interface. Writing for the tangent to the interface then this condition at the interface is described mathematically by the conditions

Steady incompressible rectilinear (unidirectional) flow.

We start with the incompressibility condition, which written explicitly, is

or

This implies

so our velocity can only be function of the and coordinates only

The non-linear term of the Navier-Stokes equation takes the form

With incompressibility and conditions killing this term, and the steady state condition 3.13 killing the term, the Navier-Stokes equation for this incompressible unidirectional steady state flow (in the absence of body forces) is reduced to

Example: Shearing flow.

The flows of this sort don’t have to be trivial. For example, even with constant pressure () as in figure (\ref{fig:continuumL11:continuumL11fig4}) we can have a “shearing flow” where the fluids at the top surface are not necessarily moving at the same rates as the fluid below that surface. We have fluid flow in the direction only, and our velocity is a function only of the coordinate.

Example: Channel flow

This time our simplified Navier-Stokes equation 3.19a is reduced to something slightly more complicated

with solution

The boundary value conditions with the coordinate system in use illustrated in figure (\ref{fig:continuumL11:continuumL11fig5}) require the velocity to be zero at the interface (the pipe walls preventing flow in the interior of the pipe)

It’s clear that this is maximized by , but we can also see this by computing

This maximum is

The flux, or flow rate is

Let’s now compute the strain () and the stress ()

stress

This can be used to compute the forces on the inner surfaces of the tube. As illustrated in figure (\ref{fig:continuumL11:continuumL11fig7}), our normals at are respectively. The traction vector in the direction is at is

FIXME: ask in class. This is the tangential force at the boundary of the wall. What is it a force on? If it is tangential, how can it act on the wall? It could act on an impediment placed right up next to the wall, but if that’s the case, why are we integrating from to ?

Disclaimer.

Problem Q1.

Statement

Solution

We need to express the relation between stress and strain in terms of Young’s modulus and Poisson’s ratio. In terms of Lam\’e parameters our model for the relations between stress and strain for an isotropic solid was given as

Computing the trace

allows us to invert the relationship

In terms of Poisson’s ratio and Young’s modulus , our Lam\’e parameters were found to be

and

Our stress strain model for the relationship for an isotropic solid becomes
we find

or

As a sanity check note that this matches (5.12) of [1], although they use a notation of instead of for Poisson’s ratio. We are now ready to tackle the problem. First we need the trace of the stress tensor

Expanding out the last bits of arithmetic the strain tensor is found to have the form

Note that this is dimensionless, unlike the stress.

Problem Q2.

Statement

Small displacement field in a material is given by

Find

\begin{enumerate}
\item the infinitesimal strain tensor ,
\item the principal strains and the corresponding principal axes at ,
\item Is the body under compression or expansion?
\end{enumerate}

Solution. infinitesimal strain tensor

Diving right in, we have

In matrix form we have

Solution. principle strains and axes

At the point the strain tensor has the value

We wish to diagonalize this, solving the characteristic equation for the eigenvalues

We find the characteristic equation to be

This doesn’t appear to lend itself easily to manual solution (there are no obvious roots to factor out). As expected, since the matrix is symmetric, a plot (\ref{fig:continuumL8:continuumProblemSet1Q2fig1}) shows that all our roots are real

with the corresponding basis (orthonormal eigenvectors), the principle axes are

Solution. Is body under compression or expansion?

To consider this question, suppose that as in the previous part, we determine a basis for which our strain tensor is diagonal with respect to that basis at a given point . We can then simplify the form of the stress tensor at that point in the object

We see that the stress tensor at this point is also necessarily diagonal if the strain is diagonal in that basis (with the implicit assumption here that we are talking about an isotropic material). Noting that the Poisson ratio is bounded according to

so if our trace is positive (as it is in this problem for all points ), then any positive principle strain value will result in a positive stress along that direction). For example at the point of the previous part of this problem (for which ), we have

We see that at this point the and components of stress is positive (expansion in those directions) regardless of the material, and provided that

(i.e. ) the material is under expansion in all directions. For the material at that point is expanding in the and directions, but under compression in the directions.

For a Mathematica notebook that visualizes this part of this problem see https://raw.github.com/peeterjoot/physicsplay/master/notes/phy454/continuumProblemSet1Q2animated.cdf. This animates the stress tensor associated with the problem, for different points and values of Poisson’s ratio , with Mathematica manipulate sliders available to alter these (as well as a zoom control to scale the graphic, keeping the orientation and scale fixed with any variation of the other parameters). This generalizes the solution of the problem (assuming I got it right for the specific point of the problem). The vectors are the orthonormal eigenvectors of the tensor, scaled by the magnitude of the eigenvectors of the stress tensor (also diagonal in the basis of the diagonalized strain tensor at the point in question). For those directions that are under expansive stress, I’ve colored the vectors blue, and for compressive directions, I’ve colored the vectors red.

This requires either a Mathematica client or the free Wolfram CDF player, either of which can run the notebook after it is saved to your computer’s hard drive.

Problem Q3.

Statement

The stress tensor at a point has components given by

Find the traction vector across an area normal to the unit vector

Can you construct a tangent vector on this plane by inspection? What are the components of the force per unit area along the normal and tangent on that surface? (hint: projection of the traction vector.)

Solution

The traction vector, the force per unit volume that holds a body in equilibrium, in coordinate form was

where was the coordinates of the normal to the surface with area . In matrix form, this is just

so our traction vector for this stress tensor and surface normal is just

We also want a vector in the plane, and can pick

or

It’s clear that either of these is normal to (the first can also be computed by normalizing , and the second with one round of Gram-Schmidt). However, neither of these vectors in the plane are particularly interesting since they are completely arbitrary. Let’s instead compute the projection and rejection of the traction vector with respect to the normal. We find for the projection

Our rejection, the component of the traction vector in the plane, is

This gives us a another vector perpendicular to the normal

Wrapping up, we find the decomposition of the traction vector in the direction of the normal and its projection onto the plane to be

The components we can read off by inspection.

Problem Q4.

Statement

The stress tensor of a body is given by

Determine the constant , , and if the body is in equilibrium.

Solution

In the absence of external forces our equilibrium condition was

In matrix form we wish to operate (to the left) with the gradient coordinate vector

So, our conditions for equilibrium will be satisfied when we have

provided , and for integer . If equilibrium is to hold along the plane, then we must either also have or also impose the restriction (for integer ).

A couple other mathematica notebooks

Some of the hand calculations done in this problem set I’ve confirmed using Mathematica. Those notebooks are available here

recall that we can decompose our force into components that refer to our direction cosines

Or in tensor form

We call this the traction vector and denote it in vector form as

Constitutive relation.

Reading: section 2, section 4 and section 5 from the text [1].

We can find the relationship between stress and strain, both analytically and experimentally, and call this the Constitutive relation. We prefer to deal with ranges of distortion that are small enough that we can make a linear approximation for this relation. In general such a linear relationship takes the form

Consider the number of components that we are talking about for various rank tensors

We have a lot of components, even for a linear relation between stress and strain. For isotropic materials we model the constitutive relation instead as

For such a modeling of the material the (measured) values and (shear modulus or modulus of rigidity) are called the Lam\’e parameters.

It will be useful to compute the trace of the stress tensor in the form of the constitutive relation for the isotropic model. We find

or

We can now also invert this, to find the trace of the strain tensor in terms of the stress tensor

Substituting back into our original relationship 3.8, and find

which finally provides an inverted expression with the strain tensor expressed in terms of the stress tensor

Special cases.

Hydrostatic compression

Hydrostatic compression is when we have no shear stress, only normal components of the stress matrix is nonzero. Strictly speaking we define Hydrostatic compression as

i.e. not only diagonal, but with all the components of the stress tensor equal.

We can write the trace of the stress tensor as

Now, from our discussion of the strain tensor recall that we found in the limit

allowing us to express the change in volume relative to the original volume in terms of the strain trace

Writing that relative volume difference as we find

or

where is called the Bulk modulus.

Uniaxial stress

Again illustrated in the plane as in figure (\ref{fig:continuumL5:continuumL5fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig2}
\caption{Uniaxial stress.}
\end{figure}

Expanding out 3.12 we have for the element of the strain tensor

or

where is Young’s modulus. Young’s modulus in the text (5.3) is given in terms of the bulk modulus . Using we find